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Question Number 187364 by ajfour last updated on 16/Feb/23

Commented by ajfour last updated on 16/Feb/23

$$Findthetimetakenbyballto$$$$reachdownthey={bx}^{2}valley.$$$$Theheight={ba}^2(\ne {ab}^2\stackrel{\bullet \bullet }{⌣})$$$$Nofriction.$$

Commented by a.lgnaoui last updated on 17/Feb/23

$$SoitReference\mathrm{\Re }(ox,oy)$$$$oy:a_y=-gv=-gt+v_{0y}$$$$at{t}_0=0y_o={ab}^2{v}_{0y}=0$$$$\Rightarrow y=-\frac{1}{2}{gt}^2+{ab}^2$$$$Alongaxis\left[ox\right):a_x=0$$$$v=\sqrt{v_x^2+v_y^2}$$$$v^2=2gx$$$$x=a=vty=0$$$$\frac{1}{2}{gt}^2={ab}^2$$$$\mathit{t}=\mathit{b}\sqrt{\frac{2\mathit{a}}{\mathit{g}}}$$$$$$

Commented by ajfour last updated on 17/Feb/23

$$Thiswontdo,wrong,ishallpost$$$$mysolution..soon(hrs..)$$

Answered by ajfour last updated on 17/Feb/23

Commented by ajfour last updated on 17/Feb/23

$$y={bx}^2$$$$\frac{dy}{dx}=2bx=\tan \theta$$$$while{\left(\frac{dy}{dt}\right)}^2=4b^2{x}^2=4by$$$$v^2=2g({ba}^2-y)$$$$v_x^2\sec {}^2\theta =2g({ba}^2-y)$$$${\left(\frac{dx}{dt}\right)}^2(1+4b^2{x}^2)=2g({ba}^2-y)$$$${\left(\frac{dy}{dt}\right)}^2\left(\frac{1+4b^2{x}^2}{4b^2{x}^2}\right)=2g({ba}^2-y)$$$$\frac{dy}{dt}\sqrt{\frac{1+4by}{4by}}=\sqrt{2g}\sqrt{{ba}^2-y}$$$$\frac{1}{\sqrt{2g}}{\int }_0^{t}dt={\int }_{{ba}^2}^0\sqrt{\frac{1}{4by}+1}\sqrt{\frac{1}{{ba}^2-y}}dy$$$$.....$$

Answered by mr W last updated on 17/Feb/23

Commented by mr W last updated on 17/Feb/23

$$ireplaceawithR,{ba}^{2}withH.$$$$eqn.ofparabolais$$$$y=H{\left(\frac{x}{R}\right)}^2$$$$let\eta =\frac{H}{R},\xi =\frac{x}{R}\in [-1,1]$$$$y^{\prime }=2H\frac{x}{R^2}=2\eta \xi$$$$y^{″}=2\frac{H}{R^2}=\frac{2\eta }{R}$$$$radiusofcurvaturer=\frac{{[1+{\left(y^{\prime }\right)}^2]}^{\frac{3}{2}}}{\mid y^{″}\mid }$$$$\Rightarrow r=\frac{{(1+4{\eta }^2\xi )}^{\frac{3}{2}}R}{2\eta }$$$$\frac{1}{2}{mv}^2=mg(H-y)=mg(H-H{\xi }^2)$$$$\Rightarrow v^2=2gH(1-{\xi }^2)$$$$\Rightarrow v=\sqrt{2gH(1-{\xi }^2)}$$$$$$$$ds=\sqrt{1+{\left(y^{\prime }\right)}^2}dx=R\sqrt{1+4{\eta }^2{\xi }^2}d\xi$$$$dt=\frac{ds}{v}=\frac{R}{\sqrt{2gH}}\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$thetimetheballtakestomovedown$$$$thevalley:$$$$T=\frac{R}{\sqrt{2gH}}{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$T=\sqrt{\frac{2H}{g}}\times \frac{1}{2\eta }{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$=k\sqrt{\frac{2H}{g}}$$$$withk=\frac{1}{2\eta }{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$=\frac{1}{2\eta }E(\frac{\pi }{2}\mid -4{\eta }^2)$$$$\sqrt{\frac{2H}{g}}isthetimefortheballto$$$$fallfreelyfromheightH.$$$$$$$$exactformulaforkisnotpossible.$$$$examples:$$$$\eta =\frac{H}{R}=1\Rightarrow k\approx 1.3176$$$$\eta =\frac{H}{R}=2\Rightarrow k\approx 1.1018$$$$\eta =\frac{H}{R}=4\Rightarrow k\approx 1.0309$$$$\eta =\frac{H}{R}=10\Rightarrow k\approx 1.0061$$

Commented by mr W last updated on 17/Feb/23

Commented by mr W last updated on 17/Feb/23

$$whentheballtakestheshortest$$$$way\left(blueinclinedplane\right):$$$$T=\sqrt{1+\frac{1}{{\eta }^2}}\sqrt{\frac{2H}{g}}$$$$\eta =1,k=1.4142$$$$\eta =2,k=1.1180$$$$\eta =4,k=1.0308$$$$\eta =10,k=1.0050$$$$thatmeanstheballmayneedmore$$$$timewhenfollowingtheshortest$$$$straightwaythanfollowingthe$$$$parabola.$$

Commented by ajfour last updated on 17/Feb/23

$$Thankyousir,extensiveanalysis!$$

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